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anygoal [31]
3 years ago
14

A triangle has side lengths measuring 20 cm, 5 cm, and n cm. Which describes the possible values of na

Mathematics
1 answer:
Marizza181 [45]3 years ago
3 0
No hablo Nintendo :(
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Scorpion4ik [409]

Scale of the new blueprint: 2 inches = 2.5 feet (or 4 inches = 5 feet)

Width of the new blueprint: 14.4 inches

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Find the area of the circle. Round your answer to the nearest hundredth. 15in
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Answer is 47.25

BRAINLIST?
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HCF Of 14608 And 720
aleksandrvk [35]
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3 years ago
Read 2 more answers
8. Based on the multiplication rule for independent events, what is the probability of getting an airplane in both boxes? Explai
Furkat [3]

Answer:

P(A and B ) = \frac{1}{4}* \frac{1}{4}=\frac{1}{16}

Step-by-step explanation:

We assume the following problem: "Consider the following ways students might create their lists using the notation:

B for block, W for watch, R for ring, and A for airplane.  The first letter represents the toy found in the first box, and the second letter represents the toy found in the second  box. The first column represents getting the block in the first box, followed by each one of the other toys. The  second column represents getting the watch in the first box, followed by each one of the other toys. The third  column is developed with the ring in the first box, and the fourth column is developed with the airplane in the first  box"

And the possible outcomes are:

BB WB RB AB

BW WW RW AW

BR WR RR AR

BA WA RA AA

As we can see we have 16 possibilities.

For this case if we use the independence of events we have the following rule. If A and B are independent events then:

P(A and B) = P(A) *P(B)

Let A = Select an airplane  from the total of 4 for the first box

B= Select an airplane from the total of 4 for the second box

For this case probability of getting an airplane selected in the  first box is 1 out of 4 since we have just one outcome possible and 4 possible.

P(A) = \frac{1}{4}

And the probability of getting an airplane selected in the second box is also 1/4 since for the first box selected we have the same number of optiosn for the second box , 4.

P(B) = \frac{1}{4}

So then we have this:

P(A and B ) = \frac{1}{4} \frac{1}{4}=\frac{1}{16}

5 0
4 years ago
URGENT!! Really need help on this test
JulsSmile [24]

Answer:

Nevermind.....I was incorrect, sorry I can't help!

6 0
3 years ago
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